Topic 1 Notes
Jeremy Orloff
1
Linear and quadratic approximation
Goal: To approximate hard to compute functions by easier functions.
Reading: More details are given in the reading from the supplementary notes: http:
//math.mit.edu/suppnotes/suppnotes01-01a/01a.pdf
Vocabulary:
• Linear approximation = linearization
• Quadratic approximation
• Geometric series
• Binomial theorem
Basic idea:
If h is small then h2 is really small and h3 is really, really small.
Example 1.1. Approximate f (x) = 3 + 4x + 5x2 + 7x3 for x near 0.
The simplest approximation is the best linear approximation. For x small we can
ignore the higher powers of x:
For x small we have f (x) ≈ 3 + 4x.
Here the wavy equal sign ’≈’ is read as ’approximately equals’.
If we want a more accurate approximation we can use the best quadratic approximation:
f (x) ≈ 3 + 4x + 5x2 .
Here we kept the first two powers of x and dropped the others.
To see why these are the best approximations we turn to calculus and draw some
pictures. While we’re at it we’ll work near an arbitrary base point x = a.
1.1
Basic linear formulas
A1. f (x) ≈ f (a) + f 0 (a)(x − a)
for x ≈ a.
A2. 1/(1 − x) ≈ 1 + x
for x ≈ 0.
A3. (1 + x)r ≈ 1 + rx
for x ≈ 0.
1
1 LINEAR AND QUADRATIC APPROXIMATION
A4. sin x ≈ x
2
for x ≈ 0.
A1 is a theoretical statement valid for all f (x). A2 − 4 are statements about specific
functions. These will be some of our building blocks for more complicated functions.
To be perfectly rigorous we should say, all f (x) that have a continuous first derivative
near x = a. In 18.01 that will be all functions that aren’t a disaster at x = a, e.g.
1/(x − 1) at x = 1.
In class we will prove A2-4 using A1. We give two proofs of A1 right now.
Algebraic proof of A1
The definition of the derivative tells us that if y = f (x) then
f 0 (x) ≈
∆y
.
∆x
Rearranging this we get
∆y ≈ f 0 (x) ∆x.
Using ∆y = f (x) − f (a) and ∆x = x − a this becomes
f (x) − f (a) ≈ f 0 (x) (x − a).
One more step (moving f (a) from the left-side to the right-side gives A1:
f (x) ≈ f (a) + f 0 (x) (x − a).
Geometric proof of A1 (The tangent line approximates the graph.)
y
y = f (x)
ytangent = f (a) + f 0 (a)(x − a)
(a, f (a))
x
a
The formula for the tangent line is shown in the figure above. Geometrically we see
that near the point (a, f (a)) the blue tangent line approximates the graph. That is
y = f (x) ≈ ytangent = f (a) + f 0 (a)(x − a)
This is exactly the approximation formula A1.
1 LINEAR AND QUADRATIC APPROXIMATION
1.2
3
Examples of linear approximation
Example 1.2. Find the best linear approximation of f (x) = (1 + x)99 (1 + 3x)77 for
x ≈ 0.
answer: This example is intended to convince you that formulas A2-4 and some
algebra are often easier than trying to apply A1 directly. Since linear approximation
ignores the higher order terms we can replace each of the factors of f (x) by its linear
approximation given by A3. (More details will be given in class.)
(1+x)99 ≈ 1+99x; (1+3x)77 ≈ 1+77·3x; so f (x) ≈ (1+99x)(1+77·3x) = 1+330x+99·231x2 ≈ 1+330x.
In the last step we again dropped the higher order term 231x2 .
Example 1.3. Aproximate f (x) = 1/(1 − sin x)2 for x ≈ 0.
answer: sin x ≈ x ⇒ f (x) ≈ 1/(1 − x)2 ≈ (1 + x)2 = 1 + 2x + x2 ≈ 1 + 2x.
Example 1.4. Approximate f (x) = ex for x near 0.
answer: None of the formulas A2-4 helps here, so we use A1 directly: f 0 (x) = ex , so
f (0) = 1, f 0 (0) = 1. Thus
ex ≈ 1 + x.
ex
near x = 0.
1−x
≈ (1 + x)(1 + x) ≈ 1 + 2x.
Example 1.5. Approximate f (x) =
answer: f (x) ≈
1.3
1+x
1−x
Basic quadratic formulas
A5. f (x) ≈ f (a) + f 0 (a)(x − a) +
A6.
f 00 (a)
(x − a)2 for x ≈ a.
2
1
≈ 1 + x + x2 for x ≈ 0.
1−x
A7. (1 + x)r ≈ 1 + rx +
r(r − 1) 2
x for x ≈ 0.
2
A8. sin x ≈ x for x ≈ 0.
A9. cos x ≈ 1 −
x2
2
for x ≈ 0.
Similar to linear approximations, A5 is theoretical and can be used to prove the
explict formulas A6-9.
√
Example 1.6. Find the best quad. approximation to 1 + 3x near 0.
√
1
1
answer: 1 + 3x = (1 + 3x)1/2 = 1 + (3x) − (3x)2 .
2
8
1 LINEAR AND QUADRATIC APPROXIMATION
4
Example 1.7. Find the best quad. approximation to f (θ) = tan θ =
answer: f (θ) ≈
sin θ
, near 0.
cos θ
θ
≈ θ(1 + θ2 /2) ≈ θ.
2
1 − θ /2
More examples:
√
Example 1.8. Find the best linear approximation of a + bx in two ways. First by
using formula (A1) and second using the basic formulas and algebra.
√
answer: (i) Give the function a name: f (x) = a + bx and then find the pieces of
(A1).
√
√
b
f (0) = a; f 0 (x) = 2√a+bx
⇒ f 0 (0) = b/2 a.
√
Using (A1): f (x) ≈ a + 2√b a x, for x ≈ 0.
1/2
√
√
bx
1 bx
(ii) f (x) = a 1 +
≈ a 1+
(same as (i)).
a
2 a
Example 1.9. Find the best quadratic approximation for ex for x ≈ 0.
answer: f (0) = 1; f 0 (x) = ex ⇒ f 0 (0); f 00 (x) = ex ⇒ f 00 (0) = 1.
⇒ ex ≈ 1 + x +
x2
.
2
Example 1.10. Find the quadratic approximation for f (x) =
1
answer: Find the pieces for (A5) (here, a = ).
2
1
1
0 1
f ( ) = 2; f 0 (x) = (1−x)
f 00 (x) =
2 ⇒ f ( 2 ) = 4;
2
1
1
Using (A5): f (x) ≈ 2 + 4(x − ) + 8(x − )2 .
2
2
2
(1−x)3
1
for x ≈ 1/2.
1−x
⇒ f 00 ( 12 ) = 16.
Example 1.11. Same problem as above, finding the answer using algebra:
answer: Let y = f (x).
Let u = x − 12 , (so x ≈ 12 ⇔ u ≈ 0).
1
2
1
1
⇒ y=
=
≈ 2(1 + 2u + 4u2 ) = 2 + 4(x − ) + 8(x − )2 .
1/2 − u
1 − 2u
2
2
(The first approximation comes using (A4).)
Example 1.12. (Special relativity: example 3 in notes §A.)
Special relativity tells us that the mass m of an object moving with respect to an
inertial frame of reference is bigger than its rest mass m0 . The formula relating the
two is
√
m = m0 c/ c2 − v 2
1 LINEAR AND QUADRATIC APPROXIMATION
5
where v is the speed of the mass and c is the speed of light. What v is needed to
produce 1% increase in mass?
answer: An increase of 1% means we want m/m0 = 1.01. Here’s a case where we
need to prepare by using some algebra to find the right way to express m/m0 so we
can use our approximation formulas.
√
m
1
= c/ c2 − v 2 = (1 − (v/c)2 )−1/2 ≈ 1 + (v/c)2 .
m0
2
Let u = v/c, so we want 1.01 = 1 + 12 u2 ⇒ .02 = u2 ⇒ u ≈
mi/sec.
1
7
⇒ v ≈ c/7 ≈ 27000
Example 1.13. Suppose you have $1000 in bank at 2% continuous interest. Approximately how much money is in the bank after 1 year? After 2 years?
answer: For this we need to know that continuous interest leads to exponential
growth in your bank balance. So if we let f (t) be the balance we have
f (t) = 1000e.02t ≈ 1000(1 + .02t + (.02t)2 /2).
Plugging in t = 1 and t = 2 gives:
f (1) ≈ 1000(1 + .02 + .0002) = 1020.20 (exact: f (1) = 1020.2013).
f (2) ≈ 1000(1 + .04 + .0008) = 1040.80 (exact: f (2) = 1040.8108).
Example 1.14. Find the best quadratic approximation of f (x) = ln(1 + x) near
x = 0.
answer: f (0) = 0;
f 00 (0) = −1.
⇒ ln(1 + x) ≈ x −
1.4
f 0 (x) = 1/(1 + x) ⇒ f 0 (0) = 1;
f 00 (x) = −1/(1 + x)2 ⇒
x2
.
2
Algebraic substitution rules
1. Can substitute a linear (quadratic) approx for any factor or divisor as long the
divisor has a constant term.
2. Once you make a linear substitution you can never recover the best quadratic
approximation.
Example 1.15. (Why we need to have a constant term) In each of the following
examples the denominator has no constant term. If we don’t cancel the extra factors
of x in the numerator and denominator we get spurious results.
1.
x(1+x)
x(2+x)
6≈
x
.
2x
1 LINEAR AND QUADRATIC APPROXIMATION
2.
6
ln(1 + x)
x
6≈ = 1. Instead = ln(1+x)/x
≈ (1 − x/2)(1 − x) ≈ 1 − 3x/2.
≈ 1−x/2
ex
1+x
x
xe
x
(Note: this would be hard to do by differentiation.)
Example 1.16. (Why we can’t get the best quadratic approximation after a linear
substitution.) Consider the function
f (x) = (1 + x + x2 + x3 )(1 + 2x + 3x2 ).
Multiplying this out and keeping just terms up to order 2 we get the quadratic
approximation near 0:
f (x) ≈ (1 + x + x2 )(1 + 2x + 3x2 ) ≈ 1 + 3x + 6x2
If first made linear approximations of each factor we get:
f (x) ≈ (1 + x)(1 + 2x) = 1 + 3x + 2x2
which is not THE BEST quadratic approximation of f (x). What happened is that
by throwing away the quadratic terms in each factor they are not included in the
product the way they should be.